Introduction of Algorithm
AlgorithmsSection 1.1: A sample algorithmic problemAn algorithmic problem is specified by describing the complete set of instances it must work on and of its outputafter running on one of these instances. This distinction, between a problem and an instance of a problem, isfundamental. The algorithmic problem known as sorting is defined as follows:
[Skiena:2008:ADM:1410219]
Problem: SortingInput: A sequence of n keys, a_1, a_2, ..., a_n.Output: The reordering of the input sequence such that a'_1 <= a'_2 <= ... <= a'_{n-1} <= a'_nAn instance of sorting might be an array of strings, such as { Haskell, Emacs } or a sequence of numbers such as{ 154, 245, 1337 }.Section 1.2: Getting Started with Simple Fizz Buzz Algorithm inSwiftFor those of you that are new to programming in Swift and those of you coming from different programming bases,such as Python or Java, this article should be quite helpful. In this post, we will discuss a simple solution forimplementing swift algorithms.Fizz BuzzYou may have seen Fizz Buzz written as Fizz Buzz, FizzBuzz, or Fizz-Buzz; they're all referring to the same thing. That"thing" is the main topic of discussion today. First, what is FizzBuzz?This is a common question that comes up in job interviews.Imagine a series of a number from 1 to 10.1 2 3 4 5 6 7 8 9 10Fizz and Buzz refer to any number that's a multiple of 3 and 5 respectively. In other words, if a number is divisibleby 3, it is substituted with fizz; if a number is divisible by 5, it is substituted with buzz. If a number is simultaneouslya multiple of 3 AND 5, the number is replaced with "fizz buzz." In essence, it emulates the famous children game"fizz buzz".To work on this problem, open up Xcode to create a new playground and initialize an array like below:// for examplelet number = [1,2,3,4,5]// here 3 is fizz and 5 is buzzTo find all the fizz and buzz, we must iterate through the array and check which numbers are fizz and which arebuzz. To do this, create a for loop to iterate through the array we have initialised:for num in number {// Body and calculation goes here}After this, we can simply use the "if else" condition and module operator in swift ie - % to locate the fizz and buzzGoalKicker.com – Algorithms Notes for Professionals 2for num in number {if num % 3 == 0 { print("\(num) fizz")} else { print(num)}}Great! You can go to the debug console in Xcode playground to see the output. You will find that the "fizzes" havebeen sorted out in your array.For the Buzz part, we will use the same technique. Let's give it a try before scrolling through the article — you cancheck your results against this article once you've finished doing this.for num in number {if num % 3 == 0 { print("\(num) fizz")} else if num % 5 == 0 { print("\(num) buzz")} else { print(num)}}Check the output!It's rather straight forward — you divided the number by 3, fizz and divided the number by 5, buzz. Now, increasethe numbers in the arraylet number = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15]We increased the range of numbers from 1-10 to 1-15 in order to demonstrate the concept of a "fizz buzz." Since 15is a multiple of both 3 and 5, the number should be replaced with "fizz buzz." Try for yourself and check the answer!Here is the solution:for num in number {if num % 3 == 0 && num % 5 == 0 { print("\(num) fizz buzz")} else if num % 3 == 0 { print("\(num) fizz")} else if num % 5 == 0 { print("\(num) buzz")} else { print(num)}}Wait...it's not over though! The whole purpose of the algorithm is to customize the runtime correctly. Imagine if therange increases from 1-15 to 1-100. The compiler will check each number to determine whether it is divisible by 3or 5. It would then run through the numbers again to check if the numbers are divisible by 3 and 5. The code wouldessentially have to run through each number in the array twice — it would have to runs the numbers by 3 first andthen run it by 5. To speed up the process, we can simply tell our code to divide the numbers by 15 directly.Here is the final code:for num in number {GoalKicker.com – Algorithms Notes for Professionals 3if num % 15 == 0 { print("\(num) fizz buzz")} else if num % 3 == 0 { print("\(num) fizz")} else if num % 5 == 0 { print("\(num) buzz")} else { print(num)}}As Simple as that, you can use any language of your choice and get startedEnjoy Coding
